Kepler’s laws MCQ Quiz - Objective Question with Answer for Kepler’s laws - Download Free PDF

Calculation:

The radius of Mars' orbit around the Sun, R′ = 4R, where R is the radius of Mercury's orbit.

The Martian year T′ = 687 Earth days.

First, apply Kepler's Third Law to find the ratio of the orbital periods:

(T′ / T)² = (R′ / R)³ = (4R / R)³ = 4³ = 64

From this, we can solve for the ratio of the periods:

T′ / T = 8

This means that Mars takes 8 times longer to orbit the Sun compared to Mercury. Therefore, the length of one year on Mercury is:

T = T′ / 8 = 687 / 8 ≈ 85.88 days

Hence, one year on Mercury is approximately 86 Earth days. The nearest option is 88 days.

Correct option is: (4) 108 days

Assuming the Sun to be a solid sphere, I = (2/5) m R²

Using conservation of angular momentum, I'ω' = Iω

⇒ (2/5) m (2R)² × (2π / T') = (2/5) m R² × (2π / T)

⇒ T' = 4T = 4 × 27 = 108 days

CONCEPT:

• Kepler's Third Law: The squares of the time periods of the planets are directly proportional to the cubes of the distances of their orbits.

T2 α R3

where T is the time period and R is the mean distance of the planet from the sun.

CALCULATION:

Given that the average distance of the plant from the sun is 1.588 times that of the earth from the sun.

\( {R_{planet} \over R_{earth}}=4\) and

Tearth = 1 yr

T2 α R3

\( {T_{planet} \over T_{earth}}= ({R_{planet} \over R_{earth}})^{3/2} \)

\( {T_{planet} \over T_{earth}}= (4)^{3/2} \)

\( {T_{planet} \over T_{earth}}= (2)^{3} \)

Tplanet = 8Tearth 

Tplanet = 8 yr

So the correct answer is option 3.

Calculation:

T² ∝ R³  

\(\left(\frac{T_{m}}{T_{s}}\right)^{2}=\left(\frac{\mathrm{R}}{\mathrm{R} / 9}\right)^{3}\)

\(\frac{\mathrm{T}_{\mathrm{m}}}{\mathrm{~T}_{\mathrm{s}}}=(3)^{3}\)

⇒ \(\mathrm{T}_{\mathrm{s}}=\left(\frac{27}{27}\right)=1 \text { day }\)

Concept:

Kepler's laws describe the motion of planets under the influence of gravitational forces that follow an inverse square law, i.e., F ∝ 1/r². The two relevant laws are:

• Kepler's law of areas: The radius vector from the Sun to a planet sweeps out equal areas in equal time intervals.
• Kepler's law of periods: The square of the orbital period (T) is directly proportional to the cube of the semi-major axis (a) of the orbit: T² ∝ a³.

In the case of a force following an inverse cube law (F ∝ 1/r³), these laws may not hold, since they are derived based on the inverse square law (F ∝ 1/r²).

Analysis:

For the inverse cube law (F ∝ 1/r³), the orbital motion and the relationship between force and distance change. The law of areas will still hold, but the law of periods will not hold because the orbital period's relationship to the semi-major axis would no longer follow the T² ∝ a³ rule.

∴ The correct answer is: 1) Kepler's law of area still holds.

Concept:

Kepler's Law:

• Kepler’s laws of planetary motion are three scientific laws describing the motion of planets around the sun.

Kepler’s first law: 

• All the planets revolve around the sun in elliptical orbits having the sun at one of the foci. 

​

Kepler’s second law states:

• The radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.
• Areal velocity is constant.
• Formula, Areal velocity, \(\frac{dA}{dt}=\frac{L}{2m}\), where L = angular momentum, m = mass 
• 

Kepler’s law of periods:

• The square of the time period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis.
• Formula, \(T=2\pi \sqrt{\frac{r^3}{GM}} \) where r = radius, G = universal gravitational constant, M = mass

Calculation:

The radius of the geostationary satellite, r₁ = R + 7R = 8R

The time period on the earth, T₁ = 24h

The radius of the second satellite, r₂ = R + 3R = 4R

Here, R = radius of the earth

The relationship between the radius of the satellite and the time period is given as,

\(T=2\pi \sqrt{\frac{r^3}{GM}} \)

\(T^2 \propto r^3\)

\(\frac{T_2}{T_1}=\sqrt{(\frac{r_2}{r_1})^3}\)

\(T_2=T_1\sqrt{(\frac{r_2}{r_1})^3}\)

\(T_2=24\times \sqrt{(\frac{4R}{8R})^3}\)

\(T_2=24\times \sqrt{(\frac{1}{2})^3} = 24\times \frac{1}{2\sqrt 2} \)

\(T_2= 24\times \frac{1}{2\sqrt 2}\times \frac{\sqrt 2}{\sqrt 2} = \frac{24\sqrt 2}{4}\)

\(T_2=6\sqrt{2} \ hr\)

Hence, the time period of the second satellite is \(6\sqrt{2} \ hr\).

CONCEPT:

Kepler’s laws of planetary motion:

• The law of Orbits: Every planet moves around the sun in an elliptical orbit with the sun at one of the foci.
• The law of Area: The line joining the sun to the planet sweeps out equal areas in equal interval of time. i.e. areal velocity is constant.
• According to this law, the planet will move slowly when it is farthest from the sun and more rapidly when it is nearest to the sun. It is similar to the law of conservation of angular momentum.
• The law of periods: The square of the period of revolution (T) of any planet around the sun is directly proportional to the cube of the semi-major axis of the orbit i.e. T2 ∝ r3

CALCULATION:

Given - 1st orbital radius (R1) = R, and 2nd orbital radius (R2) = 4R 

• According to the law of periods

⇒ T2 ∝ r3

\(⇒ (\frac{T_1}{T_2})^2=(\frac{R_1}{R_2})^3\)

\(⇒ (\frac{T_1}{T_2})^2=(\frac{R}{4R})^3=\frac{1}{64}\)

\(\Rightarrow \frac{T_1}{T_2}=(\frac{1}{64})^\frac{1}{2}=\frac{1}{8}\)

CONCEPT:

• Kepler's first law: Also known as the law of ellipses explains that planets orbit the sun in a path described as an ellipse.
• Angular Momentum (L): The angular momentum (L) of a rigid object is defined as the product of the moment of inertia (I) and the angular velocity (ω).

L = I × ω or

L = MVR (in terms of linear motion)

where M is the mass of the body, V is its linear velocity, and R is its radius of motion.

• Conservation of angular momentum: If no external force or torque acts on a system, then the angular momentum of that system remains constant.

M VARA = M VBRB

where M is the mass of the body, V_A and V_B are velocities at points A and B, and R_A AND R_B are radii at these points.

CALCULATION:

Given that:

OB/OA = R = R_B/R_A

Using conservation of angular momentum:

L_A = L_B

M_E V_A R_A = M_E V_B R_B

\(\frac{V_A}{V_B}=\frac{R_B}{R_A}=R\)

So the correct answer is option 3.

The correct answer is option 3) i.e. Law of equal areas

CONCEPT:

• Kepler's laws of planetary motion
  • The first law (Law of orbits): All the planets revolve around the sun in elliptical orbits having the sun at one of the foci.
  • The second law (Law of areas): The radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.

The planet P moves around the sun in an elliptical orbit. The shaded area is the area ΔA swept out in a small interval of time Δt.

\({\rm{i}}.{\rm{e}}.{\rm{\;}}\frac{{{\rm{dA}}}}{{{\rm{dt}}}} = {\rm{constant}}\)

• Third law (Law of periods): The square of the time period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis.

​EXPLANATION:

• From the second law of Kepler's planetary motion, it is inferred that the area swept by the planet orbiting around the Sun is equal for a given duration of time.

Hence, Kepler's second law is also known as the law of equal areas.

CONCEPT:

Johannes Kepler proposed laws of planetary motion.

• Kepler’s First law: Every planet revolves around the Sun in an elliptical orbit and Sun is situated at one of its two foci. It is also termed as ‘the Law of Orbits’.
• Conservation of Angular Momentum: The velocity and distance from the Sun both change as the planet moves in an elliptical orbit, but the product of the velocity times the distance stays constant.

L = m v r,

Where m is the mass of the planet, v is the planet's orbital velocity and r is the distance can be taken as the semi-major axis of the orbit (the distance between sun and planet).

EXPLANATION:

As from the law of conservation of angular momentum; we can say that the speed of the mercury will be the maximum when its distance from the sun is the minimum;

∴ m v r = constant

So, the speed of planet mercury at a maximum at point ‘A’ because the distance between the sun and the mercury is the lowest at ‘AS’ in the figure.

CONCEPT:

Kepler’s laws of planetary motion:

• The law of Orbits/First law: Every planet moves around the sun in an elliptical orbit with the sun at one of the foci.

• The law of Area/second law: The line joining the sun to the planet sweeps out equal areas in equal interval of time i.e. areal velocity is constant.
  • According to this law, the planet will move slowly when it is farthest from the sun and more rapidly when it is nearest to the sun. It is similar to the law of conservation of angular momentum.
• The law of periods/Third law: The Square of the period of revolution (T) of any planet around the sun is directly proportional to the cube of the semi-major axis of the orbit (r). Hence T² ∝ r³.

\(\therefore Time\;period\;\left( T \right) ∝ \;{r^{\frac{3}{2}}}\)

.EXPLANATION:

• According to the law of areas, the line joins any planet to the sun sweeps equal areas in equal intervals of time.
• This law also states that planets appear to move slower when they are farther from the sun than when they are nearer.

The correct answer is option 2) i.e. n12d13 = n22d23

CONCEPT:

• Kepler's laws of planetary motion
  • The first law (Law of orbits): All the planets revolve around the sun in elliptical orbits having the sun at one of the foci.
  • The second law (Law of areas): The radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.
  • The third law (Law of periods): The square of the time period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis.

EXPLANATION:

From the third law of Kepler, the square of the time period (T) is directly proportional to the cube of the mean distance (d).

⇒ T2 ∝ d3

We also know that, time period, \(T =\frac{1}{frequency} = \frac{1}{n}\)

\(\Rightarrow\frac {1}{n^2} \propto d^3\)

For the two planets revolving around the sun, 

\(\Rightarrow\frac {n_2^2}{n_1^2} \propto \frac{d_1^3}{d_2^3}\)

\(\Rightarrow n_1^2d_1^3 = n_2^2d_2^3\)

The correct answer is option 4) i.e. Law of periods

CONCEPT:

• Kepler's laws of planetary motion
  • The first law (Law of orbits): All the planets revolve around the sun in elliptical orbits having the sun at one of the foci.
  • The second law (Law of areas): The radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.
  • The third law (Law of periods): The square of the time period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis.

​
EXPLANATION:

• Kepler's third law of planetary motion describes the time period of revolution of planets around the sun in an orbit. Hence, it is also known as the law of periods.

The correct answer is option 3) i.e. Aphelion

CONCEPT:

• Kepler's laws of planetary motion
  • The first law (Law of orbits): All the planets revolve around the sun in elliptical orbits having the sun at one of the foci.
  • The second law (Law of areas): The radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.
  • Third law (Law of periods): The square of the time period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis.

​
EXPLANATION:

• From Kepler's first law, we know that the Earth orbits the Sun in an elliptical path. The Sun is situated at one of the foci of this elliptical orbit.
• Aphelion is the point on the Earth's orbit that is farthest away from the Sun.
• Perihelion is the point on the Earth's orbit that is nearest to the Sun.

Concept: 

Angular momentum:

• Angular momentum is the property of any rotating object given by moment of inertia times angular velocity.
  • It is a vector quantity.
  • Its SI unit is kg-m2/sec.
• If I and ω are the moment of inertia and the angular velocity respectively, then the angular momentum is given as,

⇒ L = Iω

Where L = angular momentum, I = moment of inertia, and ω = angular velocity

Kepler’s 2nd Law of Areas:

• This law is known as the law of areas.
• The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.
• The rate of change of area with time will be constant. 

Explanation:

As we know from Kepler’s second law

\(\frac{Δ A}{Δ t} = constant\) (Change in area swept with respect to time is constant). This can be proved by the conservation of angular momentum. 

\(Δ A = \frac{1}{2} × r × v\ Δ t\) -- (1)

v is speed, Δ t is time passed in covering the area, so the linear distance travelled is v .Δ t

Momentum p = mv, so v = p / m

So, putting this in (1)

\(Δ A = \frac{1}{2} × r × \frac{p}{m} \ Δ t\)

\(\implies \frac{Δ A}{Δ t} = \frac{pr}{2m}\)

Angluar momentum L = Linear momentum (p) × radius (r)

For a central force, which is directed along r, L is a constant. 

So, Δ A \ Δ t is constant as L, and m both are constant. 

So, we can say that Kepler’s law of “Areal Velocity is constant” is equivalent to law of conservation of Angular Momentum. 

Additional Information

• The first law (Law of orbits): All the planets revolve around the sun in elliptical orbits having the sun at one of the foci.
• The second law (Law of areas): The radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.
• Third law (Law of periods): The square of the time period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis.